Determinant of a matrix

Every square matrix carries a hidden single number that tells you, instantly, whether it can be "undone." That number is the determinant. Feed a matrix a determinant of zero and something has collapsed inside it — a row is a repeat of another, an equation is really two copies of the same line — and no inverse exists. Feed it a non-zero determinant and the matrix is perfectly well-behaved: invertible, and the linear system it represents has exactly one solution. This article builds the idea from scratch: what a determinant is, how $2\times2$ and $3\times3$ determinants are written and expanded, and why the single test $|A| \neq 0$ decides singular versus non-singular. Everything later in the Determinants chapter — cofactors, adjoint, inverse, solving equations by matrices — is built to make use of this one number.

The core insight

A determinant is the single real number $|A|$ attached to a square matrix $A$ that is non-zero exactly when $A$ is invertible.

The core idea

A determinant exists only for a square matrix (equal rows and columns); a $2\times3$ matrix simply has none. Write the matrix with square brackets and its determinant with vertical bars — the two notations are never interchangeable, since a matrix is an array and a determinant is one number. For order 2,

A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \qquad |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc,

where: $a, b, c, d$ are the four entries of $A$ read straight off the matrix, and $ad - bc$ is the product of the leading diagonal minus the product of the other diagonal. For order 3 the determinant is built out of $2\times2$ determinants by cofactor expansion along a row (or column), alternating $+, -, +$ in sign as you move across the row:

\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix},

where: each $2\times2$ determinant multiplying $a_{1j}$ is the minor left after deleting row 1 and column $j$ from the original matrix, and the alternating signs $+,-,+$ are the row-1 slice of the checkerboard sign pattern shown below. The single fact that powers the rest of the chapter is the invertibility test:

$|A| \neq 0 \Rightarrow A$ is non-singular, and $A^{-1}$ exists.
$|A| = 0 \Rightarrow A$ is singular, and $A^{-1}$ does not exist.

A determinant is not "the matrix written more simply" — many different matrices can share one determinant value. It is a diagnostic number: a health check that tells you, before you do any further algebra, whether the matrix is well-behaved enough to invert.

Figure

A three by three grid of plus and minus signs alternating like a checkerboard, showing the cofactor sign pattern used when expanding a determinant along any row or column

Derivation: expanding a 3x3 determinant using 2x2 determinants

Derivation AExpanding a 3x3 determinant using 2x2 determinants

The goal is to show, starting only from the $2\times2$ rule $|A| = ad - bc$ that a student already knows, exactly how the $3\times3$ expansion formula is built up entry by entry along the first row, why each term needs its own smaller determinant (its minor), and why the signs alternate $+, -, +$ rather than all being positive, so that the formula stops looking like a rule to memorise and starts looking like a pattern that can be reconstructed on the spot.

Step 1. Take a general $3\times3$ matrix $A$ with entries $a_{ij}$ , row $i$ , column $j$ . To expand along row 1, look at each entry $a_{11}, a_{12}, a_{13}$ in turn and ask: if that entry's row and column were deleted, what $2\times2$ matrix is left behind? Deleting row 1 and column 1 leaves the $2\times2$ block sitting in the bottom-right, deleting row 1 and column 2 leaves the block skipping column 2, and deleting row 1 and column 3 leaves the block skipping column 3. Each of these leftover $2\times2$ determinants is called the minor of that entry.

Step 2. Multiply each entry by its own minor. This gives three products: $a_{11}\times(\text{minor of }a_{11})$ , $a_{12}\times(\text{minor of }a_{12})$ , $a_{13}\times(\text{minor of }a_{13})$ . At this stage nothing has a sign attached yet — just an entry times a smaller determinant.

Step 3. Attach alternating signs. The rule is that position $(i,j)$ gets sign $(-1)^{i+j}$ , so along row 1 the signs run $+,-,+$ for $j = 1, 2, 3$ (since $1+1=2$ is even, $1+2=3$ is odd, $1+3=4$ is even). Adding the signs to Step 2 gives exactly the formula stated above: $|A| = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}$ , where $M_{1j}$ denotes the minor of $a_{1j}$ .

Step 4. Confirm the pattern is not arbitrary by checking it reduces correctly. If the third row and column were absent (order 2 shrinking to order 1 territory conceptually), the same delete-a-row-and-column logic is exactly how the $2\times2$ formula $ad-bc$ itself can be seen as built from $1\times1$ "determinants" (single entries). The $3\times3$ rule is the same construction one level up, and the identical logic — expand along any row or any column, always with the checkerboard sign — is what "Determinant of a 3x3 matrix" develops fully with worked practice.

Worked example

Check whether $B = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 5 & 0 \\ 0 & 3 & -9 \end{bmatrix}$ is singular or non-singular, expanding along the first row.

Step 1. Write the three minors for row 1, deleting row 1 and the relevant column each time:

|B| = 1\begin{vmatrix}5 & 0\\3 & -9\end{vmatrix} - 0\begin{vmatrix}-1 & 0\\0 & -9\end{vmatrix} + 2\begin{vmatrix}-1 & 5\\0 & 3\end{vmatrix}

Step 2. Evaluate each $2\times2$ minor using $ad-bc$ : $\begin{vmatrix}5 & 0\\3 & -9\end{vmatrix} = (5)(-9)-(0)(3) = -45$ , and $\begin{vmatrix}-1 & 5\\0 & 3\end{vmatrix} = (-1)(3)-(5)(0) = -3$ (the middle minor is not even needed, since it is multiplied by $0$ ).

Step 3. Substitute back: $|B| = 1(-45) - 0(\cdots) + 2(-3) = -45 - 6 = -51$ .

Step 4. Since $|B| = -51 \neq 0$ , matrix $B$ is non-singular and $B^{-1}$ exists.

Competitive stretch

Re-check $|B|$ by expanding along column 1 instead of row 1, to see the answer does not depend on the choice of row or column. The signs down column 1 are $+, -, +$ (positions $(1,1), (2,1), (3,1)$ ), so

|B| = 1\begin{vmatrix}5 & 0\\3 & -9\end{vmatrix} - (-1)\begin{vmatrix}0 & 2\\3 & -9\end{vmatrix} + 0\begin{vmatrix}0 & 2\\5 & 0\end{vmatrix} = 1(-45) + 1(0-6) + 0 = -45 - 6 = -51.

Same answer, $|B| = -51$ , confirming $B$ is non-singular however the expansion is done.

Common traps

Writing a determinant with square brackets, or a matrix with vertical bars — the two notations mean different things and are never swapped.
Trying to find "the determinant" of a non-square matrix — determinants are defined only for square matrices.
Getting the checkerboard signs wrong, especially forgetting the middle sign of row 1 (position $(1,2)$ ) is negative, not positive.
Confusing $|A| = 0$ (a singular matrix, no inverse) with $A = 0$ (the all-zero matrix) — a non-zero matrix, like $\begin{bmatrix}2&3\\4&6\end{bmatrix}$ , can still have determinant zero.

Ask the tutor: Have a specific $2\times2$ or $3\times3$ matrix you want checked for singular versus non-singular — ask the AI tutor to walk through the expansion with you step by step.

Why do the signs alternate when expanding a determinant instead of everything just being added

The alternating $+,-,+$ pattern comes from the position rule $(-1)^{i+j}$ : it is not a decoration, it is what makes the expansion give the same correct number regardless of which row or column you choose to expand along. If every term were added with a plus sign, expanding along row 1 versus row 2 of the same matrix would give different answers, which cannot be right for something meant to be a single, well-defined number attached to the matrix.

Do I always have to expand along the first row

No. You may expand along any row or any column, and every valid choice gives the identical determinant value, as the worked example above checks directly by comparing a row-1 expansion with a column-1 expansion. The practical tip is to expand along whichever row or column has the most zeros, since a zero entry cancels its entire term without any computation, as row 1 of matrix $B$ shows.

Is a determinant of zero the same as saying the matrix itself is the zero matrix

No, and this is one of the most common mix-ups. $|A| = 0$ only says the matrix is singular — some kind of internal dependency exists among its rows or columns, such as one row being a multiple of another. The matrix itself can be full of non-zero numbers; $\begin{bmatrix}2&3\\4&6\end{bmatrix}$ has every entry non-zero yet its determinant is exactly $0$ , because row 2 is twice row 1.